Let's break down and solve the problem **step-by-step**. --- ## **Given Data** - **Flow rate, \( Q \):** \( .030 \, \text{m}^3/\text{min}) - **Convert to \( \text{m}^3/\text{s} \):** \[ Q = \frac{.030}{60} = .0005 \, \text{m}^3/\text{s} \] - **Velocity at inlet, \( V_1 \):** \( 1.5 \, \text{m/s} \) - **Diameter at inlet, \( D_1 \):** \( 1.5 \, \text{cm} = .015 \, \text{m} \) - **Velocity at outlet, \( V_2 \):** \( 5 \, \text{m/s} \) - **Diameter at outlet, \( D_2 \):** \( .5 \, \text{cm} = .005 \, \text{m} \) - **Gauge pressure at outlet, \( P_2 \):** \( 15 \, \text{psi} = 103.4 \, \text{kPa} \) - **Density of water, \( \rho \):** \( 100 \, \text{kg/m}^3 \) - **Momentum-flux correction factor, \( \beta \):** \( 1.04 \) --- ## **(a) Calculate the inlet area (\( A_1 \))** \[ A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (.015)^2 = \frac{\pi}{4} \times .000225 = .000177 \, \text{m}^2 \] --- ## **(b) Calculate the gauge pressure at the hose inlet (\( P_1 \))** Use the **Bernoulli Equation** (for horizontal hose, neglecting height difference): \[ P_1 + \frac{1}{2}\rho V_1^2 = P_2 + \frac{1}{2}\rho V_2^2 \] Solve for \( P_1 \): \[ P_1 = P_2 + \frac{1}{2}\rho (V_2^2 - V_1^2) \] Plug in values: - \( P_2 = 103.4 \times 10^3 \, \text{Pa} \) - \( \rho = 100 \, \text{kg/m}^3 \) - \( V_2 = 5 \, \text{m/s} \), \( V_1 = 1.5 \, \text{m/s} \) \[ P_1 = 103400 + \frac{1}{2} \times 100 \times (25 - 2.25) \] \[ = 103400 + 500 \times 22.75 \] \[ = 103400 + 11375 \] \[ = 114775 \, \text{Pa} = 114.8 \, \text{kPa} \] **Final Answer (b):** \[ \boxed{P_1 = 114.8 \, \text{kPa}} \] --- ## **(c) Calculate the anchoring force (magnitude and direction)** ### **Step 1: Mass flow rate** \[ \dot{m} = \rho Q = 100 \times .0005 = .5 \, \text{kg/s} \] ### **Step 2: Apply the momentum equation (control volume form):** \[ F_x = \sum (\text{momentum out}) - \sum (\text{momentum in}) + (P_1 A_1 - P_2 A_2) \] For a steady flow, horizontal hose: \[ F_x = \beta_2 \dot{m} V_2 - \beta_1 \dot{m} V_1 + (P_1 A_1 - P_2 A_2) \] Given \( \beta_1 = \beta_2 = 1.04 \): Let's calculate each term: #### **Momentum terms:** \[ \dot{m} = .5 \, \text{kg/s} \] \[ \beta_2 \dot{m} V_2 = 1.04 \times .5 \times 5 = 2.6 \] \[ \beta_1 \dot{m} V_1 = 1.04 \times .5 \times 1.5 = .78 \] #### **Pressure force terms:** \[ A_1 = .000177 \, \text{m}^2 \] \[ A_2 = \frac{\pi}{4} (.005)^2 = .0000196 \, \text{m}^2 \] \[ P_1 A_1 = 114775 \times .000177 \approx 20.3 \, \text{N} \] \[ P_2 A_2 = 103400 \times .0000196 \approx 2.03 \, \text{N} \] #### **Total anchoring force:** \[ F_x = (2.6 - .78) + (20.3 - 2.03) \] \[ = 1.82 + 18.27 \] \[ = 20.09 \, \text{N} \] **Direction:** The force acts **opposite** to the flow direction, i.e., to hold the hose in place. --- ## **Final Answers** ### **(a) Area at the inlet** \[ \boxed{A_1 = .000177 \, \text{m}^2} \] ### **(b) Gauge pressure at the hose inlet** \[ \boxed{P_1 = 114.8 \, \text{kPa}} \] ### **(c) Anchoring force required (magnitude and direction)** \[ \boxed{20.1 \, \text{N} \text{ (opposite to flow direction)}} \] --- **Let me know if you need more details on any step!**